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3 years ago

Stainless steel, tell us about its properties and what should be taken into account when using it?

Physics

1 answer:

Nuetrik [128]3 years ago

3 0

It has low maintenance Which is long lasting, and environmentally friendly which means it’s recyclable

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In the photo, a locomotive has broken through the wall of a train station. During the collision, what can be said about the forc

Pavlova-9 [17]

Answer:

Whether the force exerted by the locomotive on the wall was larger

Than the force the locomotive could exert on the wall.

Explanation:

The Newton's third law of motion States that every force have it's equal and opposite reaction force, whose magnitude is the same as the applied force. Therefore the magnitude of these opposite forces will be equal.

So we have;

F12=-F21

F12 is the force in a direction

-F21 is the force in the opposite direction.

Therefore we see that the magnitude of the force the locomotive exerts on the wall is equal to the one the wall exerts on the locomotive. Both magnitudes are equal but in opposite directions.

6 0

3 years ago

Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thru

suter [353]

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

= π/4 * ( 2.6)^2 = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

= (1900 * 10^3) / ( 8.9480)

= 212.3373 kN

8 0

3 years ago

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?

AnnZ [28]

So, this is a problem where the accleration is not provided, since it is implied. The only acceleration is acceleration due to gravity (9.8 m/s)

The equation we will use for this problem is $V^2 =V_{0}^2 + 2a (X-X_0)$

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.

We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest. We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

$V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69$

So, the ball will go 24.69 meters up

4 0

3 years ago

Florida law applies the same protections given to pedestrians to any person riding or leading an animal upon a roadway

dmitriy555 [2]

Answer:

True

Explanation:

The Florida law states that a person driving an animal-drawn vehicle has the same protections that are applicable to a person driving a vehicle and the protections that are applicable to pedestrians also apply to a person riding or leading an animal upon a roadway or the shoulder of it. According to this, the answer is that the statement is true.

5 0

3 years ago

15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea

Rudiy27

Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

V2 = 4.8L ( i.e 4.8 x 10^-3 m3)

Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

3 0

3 years ago